POJ 1679 The Unique MST（次小生成树）

最新推荐文章于 2019-05-04 17:33:50 发布

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本文介绍了一种算法来判断给定的无向图是否拥有唯一的最小生成树（MST）。通过计算最小生成树与次小生成树的总成本进行对比，以此来确定最小生成树是否唯一。

The Unique MST

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 31994		Accepted: 11572

Description

Given a connected undirected graph, tell if its minimum spanning tree is unique.

Definition 1 (Spanning Tree): Consider a connected, undirected graph G = (V, E). A spanning tree of G is a subgraph of G, say T = (V', E'), with the following properties:
1. V' = V.
2. T is connected and acyclic.

Definition 2 (Minimum Spanning Tree): Consider an edge-weighted, connected, undirected graph G = (V, E). The minimum spanning tree T = (V, E') of G is the spanning tree that has the smallest total cost. The total cost of T means the sum of the weights on all the edges in E'.

Input

The first line contains a single integer t (1 <= t <= 20), the number of test cases. Each case represents a graph. It begins with a line containing two integers n and m (1 <= n <= 100), the number of nodes and edges. Each of the following m lines contains a triple (xi, yi, wi), indicating that xi and yi are connected by an edge with weight = wi. For any two nodes, there is at most one edge connecting them.

Output

For each input, if the MST is unique, print the total cost of it, or otherwise print the string 'Not Unique!'.

Sample Input

Sample Output

3
Not Unique!

题意：

判断最小生成树唯一否。

POINT：

算出最小和次小生成树，比较一下。

不能生成最小生成树的输出0，题目里好像没给

#include <vector>
#include <stdio.h>
#include <iostream>
#include <stdio.h>
#include <queue>
#include <string.h>
using namespace std;
#define LL long long
const LL inf = 0x3f3f3f3f;
const LL maxn =111;
LL cost[maxn][maxn];
LL used[maxn][maxn];
LL n,m;
LL Max[maxn][maxn];
LL low[maxn],pre[maxn];
LL prime()
{
    memset(low,inf,sizeof low);
    memset(Max,0,sizeof Max);
    memset(used,0,sizeof used);
    LL vis[maxn];
    memset(vis,0,sizeof vis);
    for(LL i=1;i<=n;i++) low[i]=cost[1][i],pre[i]=1;
    low[1]=0;
    vis[1]=1;
    LL ans=0;
    for(LL j=1;j<n;j++)
    {
        LL minw=inf;
        LL p=-1;
        for(LL i=1;i<=n;i++)
        {
            if(!vis[i]&&minw>low[i])
            {
                minw=low[i];
                p=i;
            }
        }
        if(minw==inf) return -1;
        vis[p]=1;
        ans+=minw;
        used[p][pre[p]]=used[pre[p]][p]=1;
        for(LL i=1;i<=n;i++)
        {
            if(vis[i]&&i!=p) Max[i][p]=Max[p][i]=max(Max[i][pre[p]],low[p]);
            else
            {
                if(low[i]>cost[p][i])
                {
                    low[i]=cost[p][i];
                    pre[i]=p;
                }
            }
        }
    }
    return ans;
}
void doit()
{
    LL ans=prime();
    if(ans==-1)
    {
        printf("0\n");
    }
    else
    {
        LL ss=inf;
        for(LL i=1;i<=n;i++)
        {
            for(LL j=1;j<=n;j++)
            {
                if(cost[i][j]==inf||used[i][j]) continue;
                ss=min(ans-Max[i][j]+cost[i][j],ss);
            }
        }
        if(ss==ans)
        {
            printf("Not Unique!\n");
        }
        else
            printf("%lld\n",ans);
    }
}
int main()
{
    LL T;
    scanf("%lld",&T);
    while(T--)
    {
        memset(cost,inf,sizeof cost);
        scanf("%lld %lld",&n,&m);
        for(LL i=1;i<=m;i++)
        {
            LL u,v,w;scanf("%lld %lld %lld",&u,&v,&w);
            cost[u][v]=cost[v][u]=w;
        }
        doit();
    }
    return 0;
}